If three numbers $a,b,c\gt0$ have a geometric average of $30$ and if the numbers $ab$,$bc$ and $ca$ have an arithmetic average of $1130$, what is the harmonic average of $a$, $b$ and $c$?
For the geometric average I found that $a⋅b⋅c=30^3$ and $ab+bc+ca=3390$ for the arithmetic average... But then I'm stuck...
The arithmetic mean referred to in the question is for the three numbers $ab,bc$ and $ca$, not $a,b,c.$ Then calculate $$\frac{ab+bc+ca}{abc}.$$